MCQ
$\mathop {\lim }\limits_{x \to 1} \frac{{1 + \log x - x}}{{1 - 2x + {x^2}}} = $
- A$1$
- B$-1$
- C$0$
- ✓$ - \frac{1}{2}$
$\mathop {{\rm{lim}}}\limits_{x \to 1} \,\,\frac{{1 + \log x - x}}{{1 - 2x + {x^2}}} = \mathop {{\rm{lim}}}\limits_{x \to 1} \,\,\frac{{\frac{1}{x} - 1}}{{ - 2 + 2x}} = \mathop {{\rm{lim}}}\limits_{x \to 1} \,\,\frac{{1 - x}}{{2x(x - 1)}}$
Again applying $ L-$ Hospital’s rule,
$\mathop {{\rm{lim}}}\limits_{x \to 1} \frac{{ - 1}}{{4x - 2}} = - \frac{1}{2}$.
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